Answer:
(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.
(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.
(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.
Step-by-step explanation:
Let <em>X</em> = number of seconds in the batch.
The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The probability mass function of <em>X</em> is:
![P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7Bn%5Cchoose%20x%7Dp%5E%7Bx%7D%281-p%29%5E%7Bn-x%7D%3B%5C%20x%3D0%2C1%2C2%2C3...)
(a)
Compute the probability that only one goblet is a second among six randomly selected goblets as follows:
![P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888](https://tex.z-dn.net/?f=P%28X%3D1%29%3D%7B6%5Cchoose%201%7D0.13%5E%7B1%7D%281-0.13%29%5E%7B6-1%7D%5C%5C%3D6%5Ctimes%200.13%5Ctimes%200.4984%5C%5C%3D0.3888)
Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.
(b)
Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:
P (X ≥ 2) = 1 - P (X < 2)
![=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776](https://tex.z-dn.net/?f=%3D1-%7B6%5Cchoose%200%7D0.13%5E%7B0%7D%281-0.13%29%5E%7B6-0%7D-%7B6%5Cchoose%201%7D0.13%5E%7B1%7D%281-0.13%29%5E%7B6-1%7D%5C%5C%3D1-0.4336%2B0.3888%5C%5C%3D0.1776)
Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.
(c)
If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.
P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)
![={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453](https://tex.z-dn.net/?f=%3D%7B4%5Cchoose%200%7D0.13%5E%7B0%7D%281-0.13%29%5E%7B4-0%7D%2B%7B5%5Cchoose%201%7D0.13%5E%7B1%7D%281-0.13%29%5E%7B5-1%7D%5C%5C%3D0.5729%2B0.3724%5C%5C%3D0.9453)
Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.